The shortest path from node 1 to node 6 is either 1-2-5-6 or 1-3-5-6 both have length 8.

The digraph can also be used to solve complicated problems. Consider this problem: We wish to minimize the time an aircraft spends at an airport. The component activities can be placed in a table:

A1

Disembark passengers

1/2 hr.

A2

Unload baggage

1 hr

A3

Clean the plane

1/2 hr

A4

Take on new passengers

1 hr.

A5

Load new baggage

1 hr.

We could simply sum all the numbers of hours but some of these tasks can be done simultaneously and some operations are independent of others. A good way to proceed is to draw a model called “The Activity Analysis Digraph.”

An Activity Analysis Digraph is constructed in the following way.

1. Represent each activity by a node A1, A2, . . . An with the time required for the activity.

2. Create two additional nodes each labeled with the number zero. One representing the job’s beginning and the other the job’s end.

3. Draw a directed edge from one activity to the next only if the first activity precedes the second.

Now that we have drawn a model, the problem is to determine the shortest time for the completion of the whole job. We can proceed as follows:

1. Denote time t measured from starting point B; t = 0

2. Rephrase the problem; given an Activity Analysis Digraph for a project. What will be the shortest time at which E, the end, can be completed?

3. Add the times for all activities on the path up to but not including E (There may be more than one path from B to E).

The critical path is the path of longest time from B to E. To determine the most efficient schedule in the problem is the critical path B, A2, A4, which has length of two hours and this gives the minimum time for the whole job to be completed.

To explain the activities A1 and A2 can both be started at time zero (Passengers can disembark and luggage be taken off at the same time).

The activity A3 cannot be started until all the passengers are taken off; the activity A4 cannot be started until A3 is completed but can be made to overlap with A5; we cannot arrive at E until both A4 and A5 are completed.